Question

Two plane mirrors are inclined at an angle of ${50}^{°}$. A ray is an incident on one of the mirrors with a certain angle of incidence. The reflected ray from this mirror is incident on the second mirror such that it gets reflected parallel to the first mirror. Find the angle of incidence at the two mirrors. Also, show that the sum of angles of incidence on the first and second mirrors is equal to the angle of inclination between the two mirrors.

Answer 1

Step 1: Ray diagram:

In the given diagram $AB$ is parallel to $ED$

$FG$ is parallel to $BD$

We know the angle of incidence $=$ angle of reflection

The angle of incidence of mirror $1$ is $r$

The angle of reflection of mirror $1$ is $r$

The angle of incidence of mirror $2$ is $\theta$

The angle of reflection in mirror $2$ is $\theta$

Step 2: Calculating the unknown angle from the ray diagram:

As $FG$ is parallel to $BD$, $FD$ acts as transversal

$\Rightarrow 2r=90-\theta$………$\left(1\right)$

and

$\angle ABD=\angle AFG={50}^{°}$ (Corresponding angles)

$\angle AFK={90}^{°}$ {$FK$ is normal}

$$\begin{gathered} \Rightarrow \angle GFK=\angle AFK-\angle AFG \\ \Rightarrow r={40}^{°}\{90-50=40\} \end{gathered}$$

Substituting the above value in $\left(1\right)$

$$\begin{gathered} \Rightarrow 2\left(40\right)=90-\theta \\ \Rightarrow 80=90-\theta \\ \Rightarrow \theta ={10}^{°} \end{gathered}$$

Step 3: Calculating the required angles:

The Sum of angles of incidence of two mirrors is $\theta +r$

$\Rightarrow 10+40$

$\Rightarrow {50}^{°}=$ Angle of inclination between the two mirrors

Hence, the angle of incidence of plane mirror 1 is ${10}^{°}$and the angle of incidence of plane mirror 2 is ${40}^{°}$ and the sum of angles of incidence of two mirrors is equal to the angle of inclination between the two mirrors.